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Chapter 1: Problem 57

Use a graphing utility to graph the function and approximate (to two decimal places) any relative minimum or relative maximum values. $$ f(x)=(x-4)(x+2) $$

Short Answer

Expert verified
The function \(f(x) = (x-4)(x+2)\) has a minimum value at (1, -9)

Step by step solution

01

Rewriting the function

Rewrite the function in the form of \(y=a(x-h)^2 + k\). Here, \(h\) and \(k\) are the x and y coordinates of the vertex. In this equation, expand the given function. It becomes \(f(x) = x^2 - 4x +2x - 8 = x^2 - 2x - 8\).
02

Find the Vertex

For a quadratic equation of the form \(y = ax^2 + bx + c\), the x-coordinate of the vertex can be found using the formula \(h = -b/2a\). Substituting the coefficients from the equation \(f(x) = x^2 - 2x - 8\), we find that \(h=2/2=1\). Substituting \(x = 1\) in to the equation we get \(y = 1 - 2 - 8 = -9\). Thus the vertex is at (1, -9).
03

Determine the Maximum or Minimum Value.

The coefficient of the x-squared term in \(f(x) = x^2 - 2x - 8\) is \(1\), a positive number. Thus, the parabola opens upwards and the vertex represents the minimum value of the function. Specifically, \(f(x) \) has a minimum value of -9 at \(x = 1\). Plotting this and checking with a graphing utility will confirm this.

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